Chapter 17:  Acids and Bases

 

Section 17-1: Arrhenius Theory, pH, and pOH

Section 17-2: Monoprotic Strong Acids and Strong Bases

Section 17-3: Monoprotic Weak Acids

Section 17-4: Polyprotic Weak Acids and Sulfuric Acid

Section 17-5: Bronsted Theory

Section 17-6: Weak Bases

Section 17-7: Lewis Structures and Acid Strength

Section 17-8: Hydrolysis Aqueous Ions as Acids and Bases

Section 17-9: Reactions of Nonmetal Oxides with Water to Create Oxoacids

Section 17-10: Experiment Determining the Molar Mass of an Unknown Monoprotic Weak Acid by Titration

Section 17-11:  pH of Buffer Solutions

Section 17-12:  pH and Titration Curves

Section 17-13:  Acid-Base Indicators

Chapter 17 Practice Exercises and Review Quizzes

 

 

 

 

Section 17-1:  Arrhenius Theory, pH, and pOH


Arrhenius theory essentially states that an acid ionizes in water to produce H+, known as a hydrogen ion or a proton, while a base ionizes in water to produce OH-, hydroxide ion.  One way to express the level of acidity or basicity in an aqueous solution is to indicate the molarity of H+, [H+], or the molarity of OH-, [OH-].  Alternatively, pH and pOH (both expressed without a unit) can be used:

 

Equation 1:  pH = - log [H+]

 

Equation 2:  pOH = - log [OH-]

 

[Note:  When rounding the result y in a calculation y = - log x, the number of decimal places in y should equal the number of significant figures in x.]

 

In any aqueous solution, the equilibrium reaction known as the autoionization of water occurs:

 

H2O (l) H+ (aq) + OH- (aq)


The equilibrium constant Kc for the reaction above is typically replaced by Kw in the equilibrium expression:

 

Equation 3:  Kw = [H+][OH-]

 

At 25C, Kw = 1.0 x 10-14.  Therefore, we can derive an equation relating pH and pOH at 25C as follows:

 

- log (Kw) = -log ([H+][OH-])

- log (1.0 x 10-14) = (- log [H+]) + (-log [OH-])

Equation 4:  14.00 = pH + pOH

 

On the pH scale at 25C:

 

pH < 7 = acidic

pH 7 = neutral

pH > 7 = basic

 

In this chapter, unless otherwise indicated, assume all solutions are aqueous and at 25C.

 

 

 

 

Section 17-2:  Monoprotic Strong Acids and Strong Bases


If an acid molecule is only capable of releasing one H+ (has only one ionizable proton), the acid is said to be monoprotic.  We can assume that all molecules of the monoprotic strong acids HCl, HBr, HI, HNO3, and HClO4 will ionize completely in solution to produce one proton and an anion having a 1- charge, with the reverse reaction being negligible (and, thus, we will use a single forward arrow rather that a double equilibrium arrow in the equation).  Therefore, the initial molarity (Mi) of the strong acid will become the molarity of H+ present in the solution at equilibrium, as demonstrated below on the RICE chart for a nitric acid solution:

 

R

    HNO3 (aq)           H+ (aq)         +       NO3- (aq)     

I

Mi

0

0

C

-Mi

+Mi

+Mi

E

0

Mi

Mi

 

Once we know the molarity of H+ at equilibrium, we can also calculate the molarity of hydroxide ion (which, even in an acidic solution, is produced due to the autoionization of water), pH, and pOH as follows:

 

Sample Exercise 17A:

 

A 44 mL sample of HCl gas, measured at 45°C and 711 torr, was dissolved in water to yield 18.5 mL of solution.  Calculate the molarity of hydrogen ion, the molarity of hydroxide ion, pH, and pOH in this solution.

 

Solution:

 

First, use the Ideal Gas Law to calculate the moles of HCl dissolved, then divide by the volume of solution to determine the initial molarity of HCl:

 

 

 

Since HCl is a strong acid, we now complete a RICE chart showing complete ionization similar to that for HNO3 above:

 

R

      HCl (aq)            H+ (aq)        +         Cl- (aq)     

I

0.085

0

0

C

-0.085

+0.085

+0.085

E

0

0.085

0.085

 

We can now complete the calculations using the equations above:

 

 

(Note that we could have also calculated pOH using Equation 2.)

 

Soluble metal hydroxides are strong bases that will ionize completely in water.  If there is only one hydroxide ion per formula unit, the initial molarity of the metal hydroxide will become the molarity of hydroxide ion present in the solution at equilibrium, as demonstrated below on the RICE chart for a sodium hydroxide solution:

 

R

     NaOH (aq)            Na+ (aq)      +         OH- (aq)     

I

Mi

0

0

C

-Mi

+Mi

+Mi

E

0

Mi

Mi

 

However, if there are two hydroxide ions per formula unit, the molarity of hydroxide ion present in the solution at equilibrium will be twice the initial molarity of the metal hydroxide:

 

Sample Exercise 17B:


A 0.0070 gram sample of calcium hydroxide was dissolved in water to create 64 mL of solution.  Calculate the molarity of hydroxide ion, the molarity of hydrogen ion, pOH, and pH in this solution.

 

Solution:

 

First, convert the mass of calcium hydroxide to moles, then divide by the volume of solution to determine the initial molarity of calcium hydroxide:

 

 

Since calcium hydroxide is a strong base, we now complete a RICE chart showing complete ionization similar to that for NaOH above:

 

R

  Ca(OH)2 (aq      Ca2+ (aq)       +     2 OH- (aq)     

I

0.0015

0

0

C

-0.0015

+0.0015

+2(0.0015)

E

0

0.0015

0.0030

 

 

 

 

 

Section 17-3:  Monoprotic Weak Acids

 

For the remainder of this chapter, assume that any acid that is not included on the strong acid list of HCl, HBr, HI, HNO3, or HClO4 is a weak acid, meaning that only a fraction of the acid molecules ionize in water.  For the ionization of a monoprotic weak acid, the RICE chart and pH calculation will differ from that of a monoprotic strong acid in the following ways:

 

1. A double equilibrium arrow will be used in the ionization equation to indicate that the reverse reaction is significant.

 

2. The initial molarity of the weak acid will not decrease to zero as the reaction proceeds.  As such, the molarity of H+ present at equilibrium will be less than the initial molarity of the weak acid.

 

3. To determine the magnitude of the change (C) line on the RICE chart, we will need to do an equilibrium calculation.  This will require knowing the value of Kc for the ionization reaction, which is typically expressed as Ka, known as the acid ionization constant.

 

The RICE chart and equilibrium calculation for all monoprotic weak acids will follow the format below:

 

R

        HA (aq)       ⇌        H+ (aq)         +          A- (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

(Note that the anion A- may contain more hydrogen atoms, but we can assume that these are not ionizable.)

 

 

 

Once we know the value of x, we can also calculate the percentage of the original sample of acid molecules that have been ionized as follows:

 

 

When comparing different monoprotic weak acids with the same initial molarity:

 

larger Ka = larger x = larger % ionization = larger [H+] = lower pH = more acidic = stronger acid

   

Sample Exercise 17C:


Write the acid ionization equation and calculate the pH and percent ionization of 0.16 M hydrofluoric acid, HF (Ka = 6.8 x 10-4).

 

Solution:

 

We complete the RICE chart and equilibrium calculation for a monoprotic weak acid, then use the value of x = [H+] to calculate the pH and percent ionization:

 

R

       HF (aq)        ⇌          H+ (aq)        +         F- (aq)     

I

0.16

0

0

C

-x

+x

+x

E

0.16 - x

x

x

 

 

 

 

If we know the pH of a solution, we can calculate [H+] as follows:

 

Equation 1a:  [H+] = 10-pH 

 

If we know the initial molarity and the pH of a monoprotic weak acid solution, we can calculate Ka as follows:

 

 Sample Exercise 17D:


A 0.58 M acetic acid solution has a pH of 2.49.  Write the acid ionization equation and calculate Ka for acetic acid, HC2H3O2 (a more condensed version of the formula CH3COOH showing the one ionizable proton at the beginning, separate from the other non-ionizable hydrogens). 

 

Solution:

 

R

  HC2H3O2 (aq)  ⇌      H+ (aq)        +   C2H3O2- (aq)     

I

0.58

0

0

C

-x

+x

+x

E

0.58 - x

x

x

 

 

 

If we know the initial molarity and the percent ionization for a monoprotic weak acid, we can calculate pH and Ka, as demonstrated at the end of the chapter in Practice Exercise 17-5.  If we know the pH and Ka for a monoprotic weak acid solution, we can calculate the initial molarity, as demonstrated in Practice Exercise 17-6.

 

Relative strengths of weak acids can also be compared using pKa values:

 

pKa = -log Ka

Ka = 10-pKa

lower pKa = larger Ka = stronger acid

 

 

 

Section 17-4:  Polyprotic Weak Acids and Sulfuric Acid


A polyprotic weak acid has more than one ionizable proton and will ionize in a stepwise series of reactions, each of which has its own distinct Ka value. 

 

A diprotic weak acid will ionize in two steps as follows:

 

1st ionization:  H2A (aq) H+ (aq) + HA- (aq)     Kc = Ka1

 

2nd ionization:  HA- (aq) H+ (aq) + A2- (aq)     Kc = Ka2

 

A triprotic weak acid will ionize in three steps as follows:

 

1st ionization:  H3A (aq) H+ (aq) + H2A- (aq)     Kc = Ka1

 

2nd ionization:  H2A- (aq) H+ (aq) + HA2- (aq)     Kc = Ka2

 

3rd ionization:  HA2- (aq) H+ (aq) + A3- (aq)     Kc = Ka3

 

The Ka values will decrease with each successive ionization:  Ka1 > Ka2 > Ka3.  In general, only the 1st ionization will produce enough H+ to affect the pH, so we will only use the RICE chart and equilibrium calculation from the 1st ionization to calculate the pH:   

 

Sample Exercise 17E:

 

Write the stepwise acid ionization equations and calculate the pH of 1.6 M phosphoric acid, H3PO4, which has the following acid ionization constants:

 

Ka1 = 7.5 x 10-3

Ka2 = 6.2 x 10-8

Ka3 = 4.2 x 10-13

 

Solution:

 

We obtain the pH by completing the RICE chart and equilibrium calculation for the 1st ionization only, after which we proceed with writing the 2nd and 3rd ionization equations:

 

R

    H3PO4 (aq)         H+ (aq)      +       H2PO4- (aq)     

I

1.6

0

0

C

-x

+x

+x

E

1.6 - x

x

x

 

 

 

2nd ionization:  H2PO4- (aq) H+ (aq) + HPO42- (aq)    

 

3rd ionization:  HPO42- (aq) H+ (aq) + PO43- (aq)  

 

 

Sulfuric acid is a unique case where the 1st ionization of H2SO4 occurs completely with a negligible reverse reaction (similar to a monoprotic strong acid), whereas only a fraction of the HSO4- anions further ionize in the 2nd ionization (similar to a diprotic weak acid):

 

1st ionization:  H2SO4 (aq) H+ (aq) + HSO4- (aq)

 

2nd ionization:  HSO4- (aq) H+ (aq) + SO42- (aq)    

 

 

Section 17-5:  Bronsted Theory


Because ammonia, NH3, does not contain hydroxide ion, Arrhenius theory does not explain how ammonia can function as a base in water. On the other hand, Bronsted theory can explain how a solution of ammonia can be basic by defining an acid and base as follows:

 

Bronsted acid = proton (H+) donor

Bronsted base = proton (H+) acceptor

 

In the case of an ammonia solution, water will act as a Bronsted acid to donate a proton to the Bronsted base ammonia.  Since the Bronsted acid water loses a proton, the formula will decrease by one hydrogen and the charge will decrease by 1, thus producing the hydroxide ion that makes the solution basic.  At the same time, the Bronsted base ammonia will gain a proton, so the formula will increase by one hydrogen and the charge will increase by 1 to produce the ammonium ion:

 

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) 

 

Although we have used Arrhenius theory to write acid ionization equations and will continue to do so, we could have chosen to use Bronsted theory instead.  In the case of solution containing the monoprotic weak acid HA, water can act as a Bronsted base to accept a proton from the Bronsted acid HA:

 

HA (aq) + H2O (l) H3O+ (aq) + A- (aq) 

 

The cation H3O+ is known as the hydronium ion and can replace H+ when calculating pH after writing the acid ionization equation according to Bronsted theory:  pH = - log [H3O+].  Since liquid water will be omitted from the RICE chart and Ka expression, the pH calculated using Bronsted theory will be the same as the pH calculated using Arrhenius theory.

 

As shown above, water acts as a Bronsted acid in an ammonia solution, but a Bronsted base in an acid solution.  A molecule or ion that can act as both a Bronsted acid or a Bronsted base in different situations is said to be amphoteric (or amphiprotic).

 

Bronsted theory also applies to proton transfer reactions in which water is not a reactant or a product:

 

Sample Exercise 17F:

 

Identify the Bronsted acids and bases in the forward and reverse directions for the reaction below:

 

CO32- (aq) + HC2O4- (aq) HCO3- (aq) + C2O42- (aq) 

 

Solution:

 

forward reaction:  Bronsted acid = HC2O4- (donates proton), Bronsted base = CO32- (accepts proton)

reverse reaction:  Bronsted acid = HCO3-, Bronsted base = C2O42-

 

 

After a proton has been donated by a Bronsted acid, the resulting molecule or ion is known as the conjugate base.  After a proton has been accepted by a Bronsted base, the resulting molecule or ion is known as the conjugate acid:  

 

Sample Exercise 17G:

 

Write the formula for:

 

a. the conjugate acid of HS-

b. the conjugate base of HC4H4O6-

 

Solution:

 

a. conjugate acid = add H+ to formula = H2S

b. conjugate base = remove H+ from formula = C4H4O62-

 

 

Section 17-6:  Weak Bases

 

For a solution of a neutral weak base B, the base ionization equation can be represented: 

 

B (aq) + H2O (l) BH+ (aq) + OH- (aq)

(proton is typically added at end of base formula)

 

Kc for the reaction above can be expressed as Kb, known as the base ionization constant.  To calculate the pH and percent ionization in a weak base solution, we complete a RICE chart and equilibrium calculation similar to that for a monoprotic weak acid.  However, the column under liquid water will be omitted from the RICE chart and water will be omitted from the Kb expression.  Also note that the calculated value of x = [OH-], not the molarity of hydrogen ions:

 

R

        B (aq)        +          H2O(l)                  BH+ (aq)        +         OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

When comparing different weak bases with the same initial molarity:

 

larger Kb = larger x = larger % ionization = larger [OH-] = higher pH = more basic = stronger base

 

Sample Exercise 17H:

 

Calculate the pH and percent ionization of 0.029 M ammonia (Kb = 1.8 x 10-5).

 

Solution:

 

R

       NH3 (aq)      +        H2O(l)                  NH4+ (aq)      +        OH- (aq)

I

0.029

 

0

0

C

-x

+x

+x

E

0.029- x

x

x

 

 

 

 

If we know the pOH of a solution, we can calculate [OH-] as follows:

 

Equation 2a:  [OH-] = 10-pOH 

 

If we know the initial molarity and the pH of a weak base solution, we can calculate Kb as follows:

 

Sample Exercise 17I:

 

A 0.026 M methylamine solution has a pH of 11.51.  Write the base ionization equation and calculate Kb for methylamine, CH3NH2.

 

 

Solution:

 

R

   CH3NH2 (aq)    +         H2O(l)              CH3NH3+ (aq)    +       OH- (aq)

I

0.026

 

0

0

C

-x

+x

+x

E

0.026- x

x

x

 

 

 

 

If we know the initial molarity and the percent ionization for a weak base, we can calculate pH and Kb, as demonstrated in Practice Exercise 17-12.  If we know the pH and Kb for a weak base solution, we can calculate the initial molarity, as demonstrated in Practice Exercise 17-13.

 

Relative strengths of weak bases can also be compared using pKb values:

 

pKb = -log Kb

Kb = 10-pKb

lower pKb = larger Kb = stronger base

 

 


Section 17-7:  Lewis Structures and Acid Strength

 

The Lewis structure for an oxoacid (or oxyacid) with the formula HmXOn will generally show the center atom X bonded to each oxygen atom, and each hydrogen atom will be bonded to a different oxygen atom, as demonstrated below for sulfurous acid, H2SO3:

 

 

(other resonance structures also acceptable)

 

To compare the relative strengths of oxoacids, use the following guidelines:

 

1. More terminal oxygens (= oxygens not bonded to a hydrogen = n-m) = oxoacid donates proton more readily = stronger acid.

2. Higher electronegativity of center atom X = oxoacid donates proton more readily = stronger acid.

 

Sample Exercise 17J:

 

Draw Lewis structures for nitric acid, HNO3, and phosphoric acid, H3PO4.  Which is the stronger acid?  Give two reasons to justify your answer.

 

Solution:

 

 

 

 

Nitric acid is stronger acid because:

 

1. HNO3 has more terminal oxygens (3-1=2) than H3PO4 (4-3=1).

2. Electronegativity for central atom N is higher than for central atom P.

 

 

The Lewis structure for a monoprotic organic acid with the formula RCOOH, where R can either represent a single atom or a group of atoms, is shown below:

 

 

The ionizable proton will be the one single-bonded to the oxygen atom on the right end of the Lewis structure, not in R.  To compare the relative strengths of organic acids, use the following general guidelines:

 

1. Higher electronegativity of atoms in R = organic acid donates proton more readily = stronger acid.

2. Highly electronegative atoms in R are closer in proximity to ionizable hydrogen = organic acid donates proton more readily = stronger acid.

 

Sample Exercise 17K:


Which of the two acids shown below is the stronger acid?  Give two reasons to justify your answer.

 

 

 

 

Solution:


The bottom acid is stronger because:

 

1. Electronegativity of F is higher than Br.

2. F is closer to ionizable proton than Br.

 


Section 17-8:  Hydrolysis – Aqueous Ions as Acids and Bases


When an ionic compound dissolves in water, it is sometimes possible for the ions in solution to function as acids or bases in a process known as hydrolysis.  Use the following guidelines to decide if an ion will hydrolyze and, therefore, affect the pH:

 

1. A cation BH+ containing an ionizable proton (ammonium, NH4+, for example) will generally hydrolyze as a weak acid:

 

BH+ (aq) H+ (aq) + B (aq)

 

We can determine Ka for the reaction above by utilizing the fact that when two reactions are added to obtain a third reaction, the product of the equilibrium constants for the two added reactions will equal the equilibrium constant for the third reaction:

 

BH+ (aq) H+ (aq) + B (aq)     Ka for BH+ = ???

B (aq) + H2O (l) BH+ (aq) + OH- (aq)     Kb for B = given in problem or found on Kb data table

__________________________________________________________________________________________

H2O (l) H+ (aq) + OH- (aq)     Kw = 1.0 x 10-14

 

(Ka for BH+)(Kb for B) = Kw

 

 

(so stronger base B = weaker conjugate acid BH+)

 

2. Metal cations will generally not hydrolyze and, therefore, will be spectator ions that do not affect the pH.  Notable exceptions include Al3+ and Fe3+.

 

3. An anion will generally hydrolyze as a weak base:

 

A- (aq) + H2O (l) HA (aq) + OH- (aq)

 

We can determine Kb for the reaction above as follows:

 

A- (aq) + H2O (l) HA (aq) + OH- (aq)     Kb for A- = ???

HA (aq) H+ (aq) + A- (aq)     Ka for HA = given in problem or found on Ka data table

_________________________________________________________________________________

H2O (l) H+ (aq) + OH- (aq)     Kw = 1.0 x 10-14

 

(Kb for A-)(Ka for HA) = Kw

 

 

(so stronger acid HA = weaker conjugate base A-)

 

4. Anions found in monoprotic strong acids (Cl-, Br-, I-, NO3-, ClO4-) do not hydrolyze as weak bases to any significant extent, so these five anions will be spectator ions that do not affect the pH.

 

Using the guidelines above, we can predict if an ionic compound solution will be acidic, basic, or neutral as follows:

 

A. If both the cation and the anion are spectator ions, the solution will be neutral.

 

B. If only the cation hydrolyzes and the anion is a spectator ion, the solution will be acidic.

 

C. If the cation is a spectator ion and only the anion hydrolyzes, the solution will be basic.

 

D. If both the cation and anion hydrolyze, then we must compare the equilibrium constants for the two hydrolysis reactions:

 

if Ka for BH+ > Kb for A-, then solution is acidic

if Ka for BH+ < Kb for A-, then solution is basic

if Ka for BH+ = Kb for A-, then solution is neutral

 

Sample Exercise 17L:


Predict whether a solution of each compound below will be acidic, basic, or neutral.  For solutions that are not neutral, show all relevant hydrolysis reactions that affect the pH and also calculate the equilibrium constant for each reaction you write using information from the following data tables:

 

Acid

Ka

HC2H3O2

1.8 x 10-5

HNO2

4.5 x 10-4

 

Base

Kb

C5H5N

1.7 x 10-9

CH3NH2

4.4 x 10-4

 

a. Mg(NO3)2

b. CH3NH3Cl [composed of CH3NH3+ and Cl-]

c. NaC2H3O2

d. C5H5NHNO2 [composed of C5H5NH+ and NO2-]

 

Solution:


a. Mg2+ and NO3- = spectator ions = solution is neutral

 

b. Cl- = spectator ion, CH3NH3+ hydrolyzes as weak acid = solution is acidic:

 

CH3NH3+ (aq) H+ (aq) + CH3NH2 (aq) 

 

 

c. Na+ = spectator ion, C2H3O2- hydrolyzes as a weak base = solution is basic:

 

C2H3O2- (aq) + H2O (l) HC2H3O2 (aq) + OH- (aq)

 

 

d. Both ions hydrolyze, so we must compare equilibrium constants:

 

C5H5NH+ (aq) H+ (aq) + C5H5N (aq) 

 

 

NO2- (aq) + H2O (l) HNO2 (aq) + OH- (aq)

 

 

Ka > Kb = solution is acidic

 

 

To calculate the pH of a solution where only the cation hydrolyzes as a weak acid and the anion is a spectator ion, we complete a RICE chart and equilibrium calculation similar to that completed earlier for a neutral monoprotic weak acid.  To calculate the pH of a solution where only the anion hydrolyzes as a weak base and the cation is a spectator ion, we complete a RICE chart and equilibrium calculation similar to that completed earlier for a neutral weak base.

 

Sample Exercise 17M:

 

For each solution below, show any relevant hydrolysis reactions and calculate the pH.

 

a. 1.6 M NH4I (Kb for NH3 = 1.8 x 10-5)

b. 0.84 M KClO (Ka for HClO = 3.0 x 10-8)

 

Solution:


a. I- = spectator ion, NH4+ hydrolyzes as a weak acid:

 

R

     NH4+ (aq)     ⇌       H+ (aq)         +      NH3 (aq)     

I

1.6

0

0

C

-x

+x

+x

E

1.6 - x

x

x

 

 

 

b. K+ = spectator ion, ClO- hydrolyzes as a weak base:

 

R

       ClO- (aq)      +        H2O(l)                 HClO (aq)       +        OH- (aq)

I

0.84

 

0

0

C

-x

+x

+x

E

0.84- x

x

x

 

 

 

 

 

If a solution contains metal cations that are spectator ions along with anions of the type HA-, a portion of the HA- ions will ionize as weak acids while the remainder of the HA- ions will hydrolyze as weak bases:

 

weak acid ionization reaction: 

HA- (aq) H+ (aq) + A2- (aq)     Ka for HA- = Ka2 for H2A


weak base hydrolysis reaction:

HA- (aq) + H2O (l) H2A (aq) + OH- (aq)     Kb for HA- = Kw/(Ka1 for H2A)

 

For anions of the type H2A-, the relevant reactions that affect the pH are:

 

weak acid ionization reaction: 

H2A- (aq) H+ (aq) + HA2- (aq)     Ka for H2A- = Ka2 for H3A


weak base hydrolysis reaction:

H2A- (aq) + H2O (l) H3A (aq) + OH- (aq)     Kb for H2A- = Kw/(Ka1 for H3A)

 

For anions of the type HA2-, the relevant reactions that affect the pH are:

 

weak acid ionization reaction: 

HA2- (aq) H+ (aq) + A3- (aq)     Ka for HA2- = Ka3 for H3A


weak base hydrolysis reaction:

HA2- (aq) + H2O (l) H2A- (aq) + OH- (aq)     Kb for HA2- = Kw/(Ka for H2A- = Ka2 for H3A)

 

To determine if an HA-, H2A-, or HA2- solution is acidic or basic, we must compare the equilibrium constants for the weak acid ionization reaction and the weak base hydrolysis reaction:

 

if Ka > Kb, then solution is acidic

if Ka < Kb, then solution is basic

 

Sample Exercise 17N:


Predict whether a solution of sodium bicarbonate, NaHCO3, will be acidic or basic.  Show all relevant reactions that affect the pH and also give the value of the equilibrium constant for each reaction you write.  For carbonic acid, H2CO3, Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11

 

Solution:


Na+ = spectator ion

 

weak acid ionization reaction: 

HCO3- (aq) H+ (aq) + CO32- (aq)    

Ka for HCO3- = Ka2 for H2CO3 = 5.6 x 10-11


weak base hydrolysis reaction:

HCO3- (aq) + H2O (l) H2CO3 (aq) + OH- (aq)    

Kb for HCO3- = 1.0 x 10-14/(Ka1 for H2CO3 = 4.3 x 10-7) = 2.3 x 10-8

 

Ka < Kb = solution is basic

 


Section 17-9:  Reactions of Nonmetal Oxides with Water to Create Oxoacids


Recall from Chapter 10 that basic hydroxide ion is produced when a metal oxide reacts with water.  When a nonmetal oxide reacts with water, the product is an oxoacid.  Note that this is not a redox reaction, so the atoms do not undergo a change in oxidation number as the reaction proceeds.

 

nonmetal oxide + water oxoacid with no change in oxidation numbers

 

For example, the reaction of SO2 (oxidation number of S = +4) and water will produce H2SO3 (oxidation number of S = +4) rather than H2SO4 (oxidation number of S = +6).

 

Sample Exercise 17O:

 

 

Will the reaction of N2O5 and water produce HNO2 or HNO3?  Write a balanced equation for the reaction.

 

Solution:

 

The reaction of N2O5 (oxidation number of N = +5) and water will produce HNO3 (oxidation number of N = +5) rather than HNO2 (oxidation number of N = +3).  The balanced equation will be N2O5 + H2O 2 HNO3.

 


Section 17-10:  Experiment – Determining the Molar Mass of an Unknown Monoprotic Weak Acid by Titration


Recall from Chapter 3 that we can determine the empirical formula of a compound if we know the percent composition by mass, after which we can determine the molecular formula using the experimentally determined molar mass. We can obtain the molar mass of an unknown monoprotic weak acid HA using titration as follows:

 

1. Dissolve a known mass of HA in water.

 

2. Titrate the HA solution with a metal hydroxide solution of known molarity from a buret.  Record the volume of metal hydroxide solution required to reach the equivalence point.

 

3.  Use stoichiometry from the net ionic equation HA (aq) + OH- (aq) H2O (l) + A- (aq) to calculate the moles of HA reacted, then divide the mass of HA by moles of HA to obtain the molar mass of HA.

   

Sample Exercise 17P:


a. An unknown monoprotic weak acid was found to be 54.53% carbon and 9.15% hydrogen by mass, with the remainder being oxygen.  Determine the empirical formula of the acid.

b. In a separate experiment, 6.7 grams of the acid was dissolved in 25 mL of water and then titrated with 0.85 M barium hydroxide.  The volume of base required to reach the equivalence point was 45 mL.  Calculate the molar mass and determine the molecular formula of the acid.

 

Solution:

 

a. 100% - 54.53% C - 9.15% H = 36.32%O by mass.

 

Assume one hundred grams of unknown acid:

 

 

4.540 mol C: 9.08 mol H: 2.270 mol O (divide each by 2.270)

= 2 mol C: 4 mol H: 1 mol O

Therefore, empirical formula is C2H4O.

 

b. 0.85 M Ba(OH)2 0.85 M Ba2+ and 2 x 0.85 M = 1.7 M OH-

Use stoichiometry from the net ionic equation HA (aq) + OH- (aq) H2O (l) + A- (aq) to calculate the moles of HA reacted, then divide the mass of HA by moles of HA to obtain the molar mass of HA.  Note that the volume of water used to dissolve the HA is not relevant in this calculation:

 

 

Since the acid is monoprotic, we can rewrite the molecular formula as HC4H7O2.

 

 

 

Section 17-11:  pH of Buffer Solutions


One type of buffer solution contains significant concentrations of both a weak acid HA and its conjugate base A-.  When HA and A- are mixed in solution, the presence of a significant concentration of A- reduces the ionization of HA in the reaction HA (aq) H+ (aq) + A- (aq).  This is an example of the common-ion effect.  We will generally assume that the common-ion effect reduces ionization of HA to an extent where the initial concentrations and moles of HA and A- in the solution upon mixing are effectively equal to their equilibrium concentrations and moles.  Therefore, for the ionization of HA in an HA/A- buffer solution, we can use initial rather than equilibrium concentrations and moles in the Ka expression below:

 

 

A variation of the expression above known as the Henderson-Hasselbalch equation can be used to calculate the pH of an HA/A- buffer solution directly and can derived as follows:

 

 

Sample Exercise 17Q:


Calculate the pH of a solution containing 0.46 grams of NaF in 75 mL of 0.20 M HF (Ka = 6.8 x 10-4).

 

Solution:


Na+ = spectator ion, HF/F- = buffer solution:

 

 

An HA/A- buffer solution is able to maintain a constant pH as water is added because the ratios [A-]/[HA] and nA-/nHA will not change upon dilution.  However, the solution will eventually cease functioning as an effective buffer if the volume is increased to a point where the concentrations of HA and A- are too low.

 

 

A second type of buffer solution contains significant concentrations of both a weak base B and its conjugate acid BH+.  When B and BH+ are mixed in solution, the presence of a significant concentration of BH+ reduces the ionization of B in the reaction B (aq) + H2O (l) BH+ (aq) + OH- (aq).  This is another example of the common-ion effect.  We will generally assume that the common-ion effect reduces ionization of B to an extent where the initial concentrations and moles of B and BH+ in the solution upon mixing are effectively equal to their equilibrium concentrations and moles.  Therefore, for the ionization of B in a B/BH+ buffer solution, we can use initial rather than equilibrium concentrations and moles in the Kb expression below:

 

 

A variation of the expression above can be used to calculate the pH of a B/BH+ buffer solution directly and can derived as follows:

 

 

Sample Exercise 17R:


Calculate the pH of a solution containing 3.0 grams of NH4Cl in 30. mL of 1.6 M NH3 (Kb = 1.8 x 10-5).

 

Solution:


Cl- = spectator ion, NH3/NH4+ = buffer solution:

 

A B/BH+ buffer solution is able to maintain a constant pH as water is added because the ratios [BH+]/[B] and nBH+/nB will not change upon dilution.  However, the solution will eventually cease functioning as an effective buffer if the volume is increased to a point where the concentrations of B and BH+ are too low.

 

 

Even if the amount of strong acid or strong base added to pure water is small, the pH can still change from pH = 7 by several units.  For example, if 0.001 mol H+ is added to water to create 1 L of 0.001 M H+, the new pH = -log (0.001) = 3.0.  Likewise, if 0.001 mol OH- is added to water to create 1 L of 0.001 M OH-, the new pOH = -log (0.001) = 3.0 and the new pH = 11.0.

 

On the other hand, the pH will only change slightly as small increments of strong acid or strong base are added to an HA/A- or B/BH+ buffer solution.  The H+ or OH- added will be neutralized by buffer components and the new pH of the buffer solution calculated as follows (based on 1:1 stoichiometric mole ratios and assuming the H+ or OH- added is the limiting reagent):

 

OH- added to HA/A- buffer:

 

neutralization reaction:  HA (aq) + OH- (aq) A- (aq) + H2O (l)

pH = pKa + log (nA- + nOH-)/(nHA - nOH-)

(only valid if nHA > nOH-)

 

H+ added to HA/A- buffer:

 

neutralization reaction:  A- (aq) + H+ (aq) HA (aq)

pH = pKa + log (nA- - nH+)/(nHA + nH+)

(only valid if nA- > nH+) 

 

H+ added to B/BH+ buffer:

 

neutralization reaction:  B (aq) + H+ (aq) BH+ (aq)

pOH = pKb + log (nBH+ + nH+) /(nB - nH+) = 14.00 - pH

(only valid if nB > nH+) 

 

OH- added to B/BH+ buffer:

 

neutralization reaction:  BH+ (aq) + OH- (aq) B (aq) + H2O (l)

pOH = pKb + log (nBH+ - nOH-)/(nB + nOH-) = 14.00 - pH

(only valid if nBH+ > nOH-) 

 

Sample Exercise 17S:


Write the net ionic equation for the neutralization reaction that occurs and calculate the pH when:

 

a. 0.001 mol HCl is added to the solution in Sample Exercise 17Q

b. 0.001 mol NaOH is added to the solution in Sample Exercise 17Q

c. 0.001 mol HCl is added to the solution in Sample Exercise 17R

d. 0.001 mol NaOH is added to the solution in Sample Exercise 17R

 

Solution:


a. Cl- = spectator ion

neutralization reaction:  F- (aq) + H+ (aq) HF (aq)

pH = 3.17 + log (0.011 – 0.001) mol F-/(0.015 + 0.001) mol HF = 2.97

 

b. Na+ = spectator ion

neutralization reaction:  HF (aq) + OH- (aq) F- (aq) + H2O (l)

pH = 3.17 + log (0.011 + 0.001) mol F-/(0.015 - 0.001) mol HF = 3.10

 

c. Cl- = spectator ion

neutralization reaction:  NH3 (aq) + H+ (aq) NH4+ (aq)

pOH = 4.74 + log (0.056 + 0.001) mol NH4+/(0.048 - 0.001) mol NH3 = 4.82

pH = 14.00 – 4.82 = 9.18 

 

d. Na+ = spectator ion

neutralization reaction:  NH4+ (aq) + OH- (aq) NH3 (aq) + H2O (l)

pOH = 4.74 + log (0.056 - 0.001) mol NH4+/(0.048 + 0.001) mol NH3 = 4.79

pH = 14.00 – 4.79 = 9.21

 

 

Rather than mixing HA and A- or B and BH+ to create a buffer solution, we can also create an HA/A- or B/BH+ buffer solution via a neutralization reaction if the H+ or OH- added is the limiting reagent and the other reactant is in excess.  If that is not the case, then the method of calculating the pH of the resulting solution after the neutralization reaction is complete is also described below:

 

OH- added to HA:

 

neutralization reaction:  HA (aq) + OH- (aq) A- (aq) + H2O (l)

nHA > nOH-:  HA/A- buffer created, pH = pKa + log (nA- produced/nHA excess)

nHA = nOH-:  A- solution created, pOH = 14.00 - pH calculated using RICE chart for hydrolysis of weak base A-

nHA < nOH-:  OH-/A- solution created, pOH = 14.00 – pH = -log [OH-]excess (hydrolysis of A- negligible)

 

H+ added to A-:

 

neutralization reaction:  A- (aq) + H+ (aq) HA (aq)

nA- > nH+:  HA/A- buffer created, pH = pKa + log (nA- excess/nHA produced)

nA- = nH+:  HA solution created, pH calculated using RICE chart for ionization of weak acid HA

nA- < nH+:  H+/HA solution created, pH = -log [H+]excess (ionization of HA negligible)

 

H+ added to B:

 

neutralization reaction:  B (aq) + H+ (aq) BH+ (aq)

nB > nH+:  B/BH+ buffer created, pOH = pKb + log (nBH+ produced/nB excess) = 14.00 - pH

nB = nH+:  BH+ solution created, pH calculated using RICE chart for hydrolysis of weak acid BH+

nB < nH+:  H+/BH+ solution created, pH = -log [H+]excess (hydrolysis of BH+ negligible)

 

OH- added to BH+:

 

neutralization reaction:  BH+ (aq) + OH- (aq) B (aq) + H2O (l)

nBH+> nOH-:  B/BH+ buffer created, pOH = pKb + log (nBH+ excess/nB produced) = 14.00 - pH

nBH+ = nOH-:  B solution created, pOH = 14.00 - pH calculated using RICE chart for ionization of weak base B

nBH+< nOH-:  OH-/B solution created, pOH = 14.00 – pH = -log [OH-]excess (ionization of B negligible)

 

 Sample Exercise 17T:


Write the net ionic equation for the neutralization reaction that occurs in each aqueous mixture below.  Which one of the four reactions creates a buffer solution?  For each mixture, describe the method of calculating the pH of the resulting solution after the neutralization reaction is complete.

 

a. 0.1 mol HBr + 0.1 mol CH3NH2 (a weak base) 

b. 0.3 mol HI + 0.1 mol KC3H5O3

c. 0.3 mol NH4NO3 + 0.1 mol NaOH

d. 0.1 mol HNO2 and 0.1 mol KOH

 

Solution:


a. Br- = spectator ion

neutralization reaction:  CH3NH2 (aq) + H+ (aq) CH3NH3+ (aq)

n of CH3NH2 = nH+:  CH3NH3+ solution created, pH calculated using RICE chart for hydrolysis of weak acid CH3NH3+

 

b. I- and K+ = spectator ions

neutralization reaction:  C3H5O3- (aq) + H+ (aq) HC3H5O3 (aq)

n of C3H5O3- < nH+:  H+/ HC3H5O3 solution created, pH = -log [H+]excess (ionization of HC3H5O3 negligible)

 

c. NO3- and Na+ = spectator ions

neutralization reaction:  NH4+ (aq) + OH- (aq) NH3 (aq) + H2O (l)

n of NH4+ > nOH-:  NH3/ NH4+ buffer created, pOH = pKb + log (n of NH4+ excess/n of NH3 produced) = 14.00 - pH

 

d. K+ = spectator ion

neutralization reaction:  HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

n of HNO2 = nOH-:  NO2- solution created, pOH = 14.00 - pH calculated using RICE chart for hydrolysis of weak base NO2-

 

 

  

 Section 17-12:  pH and Titration Curves


A titration curve is a graph showing the pH at different points during a titration:

 

Strong Acid + Strong Base Titration

 

When a strong acid is titrated with a strong base, the net ionic equation for the neutralization reaction is H+ (aq) + OH- (aq) H2O (l), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of base needed to reach the equivalence point as follows:

 

LH+ × MH+ = nH+ = nOH-/ MOH- = LOH-

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial:

pH = -log [H+]

 

Point B = between start of titration and equivalence point:

pH = -log [H+]excess = -log [(nH+ - nOH-)/L total]

 

Point C = equivalence point:

nH+ = nOH- = only H2O (l) present so pH = 7

 

Point D = after equivalence point:

pOH = -log [OH-]excess = -log [(nOH- - nH+)/L total] = 14.00 - pH   

 

 Sample Exercise 17U:


a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 24 mL of 0.30 M HCl with 0.20 M NaOH and calculate the volume of base needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 22 mL of base has been added

iii. the pH at the equivalence point

iv. the pH after 48 mL of base has been added

 

Solution:


a. Cl- and Na+ = spectator ions, neutralization reaction:  H+ (aq) + OH- (aq) H2O (l)

0.024 L (0.30 mol H+/1 L) = 0.0072 mol H+ = 0.0072 mol OH- (1 L/0.20 mol OH-) = 0.036 L = 36 mL base needed

 

b.

i. pH = -log (0.30) = 0.52

 

ii. 0.022 L (0.20 mol OH-/1 L) = 0.0044 mol OH- added

pH = -log [(0.0072 - 0.0044) mol H+ excess/(0.024 + 0.022) L total] = 1.22

 

iii. only H2O present, pH = 7

 

iv. 0.048 L (0.20 mol OH-/1 L) = 0.0096 mol OH- added

pOH = - log [(0.0096 - 0.0072) mol OH- excess/(0.024 + 0.048) L total] = 1.48

pH = 14.00 – 1.48 = 12.52


 

Strong Base + Strong Acid Titration

 

When a strong base is titrated with a strong acid, the net ionic equation for the neutralization reaction is OH- (aq) + H+ (aq) H2O (l), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of acid needed to reach the equivalence point as follows:

 

LOH- × MOH- = nOH- = nH+/ MH+ = LH+

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial:

pOH = -log [OH-] = 14.00 - pH

 

Point B = between start of titration and equivalence point:

pOH = -log [OH-]excess = -log [(nOH- - nH+)/L total] = 14.00 - pH

 

Point C = equivalence point:

nOH- = nH+ = only H2O (l) present so pH = 7

 

Point D = after equivalence point:

pH = -log [H+]excess = -log [(nH+ - nOH-)/L total]

 

We will demonstrate how to calculate the pH at different points in the titration of a strong base with a strong acid below in Practice Exercise 17-29.

 

 

Weak Acid + Strong Base Titration

 

When a weak acid is titrated with a strong base, the net ionic equation for the neutralization reaction is HA (aq) + OH- (aq) H2O (l) + A- (aq), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of base needed to reach the equivalence point as follows:

 

LHA × MHA = nHA = nOH-/ MOH- = LOH-

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial pH calculated using RICE chart for ionization of weak acid HA:

 

R

       HA (aq)               H+ (aq)         +         A- (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

Point B = between start of titration and equivalence point = HA/A- buffer solution created:

nOH- added = nHA reacted = nA- produced

nHA initial - nHA reacted = nHA excess

pH = pKa + log (nA- produced/nHA excess)

halfway to equivalence point:  nA- produced = nHA excess, so pH = pKa

 

Point C = equivalence point, pOH = 14.00 - pH calculated using RICE chart for hydrolysis of weak base A-:

 

R

        A- (aq)        +       H2O(l)                 HA (aq)         +        OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

pH > 7 at equivalence point

 

Point D = after equivalence point (hydrolysis of A- negligible):

pOH = -log [OH-]excess = -log [(nOH- - nHA)/L total] = 14.00 - pH   

 

Sample Exercise 17V:


a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 16 mL of 0.20 M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10 M KOH and calculate the volume of base needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 14 mL of base has been added

iii. the pH at the equivalence point

iv. the pH after 44 mL of base has been added

 

Solution:


a. K+ = spectator ion, neutralization reaction:  HC2H3O2 (aq) + OH- (aq) H2O (l) + C2H3O2- (aq)

0.016 L (0.20 mol HC2H3O2/1 L) = 0.0032 mol HC2H3O2 = 0.0032 mol OH- (1 L/0.10 mol OH-) = 0.032 L = 32 mL base needed

 

b.

i.

R

  HC2H3O2 (aq)         H+ (aq)        +     C2H3O2- (aq)     

I

0.20

0

0

C

-x

+x

+x

E

0.20 - x

x

x

 

 

ii.

0.014 L (0.10 mol OH-/1 L) = 0.0014 mol OH- added = 0.0014 mol HC2H3O2 reacted = 0.0014 mol C2H3O2- produced

0.0032 mol HC2H3O2 initial - 0.0014 mol HC2H3O2 reacted = 0.0018 mol HC2H3O2 excess

pKa for HC2H3O2  = -log (1.8 x 10-5) = 4.74

pH = 4.74 + log (0.0014 mol C2H3O2-/0.0018 mol HC2H3O2) = 4.63

 

iii.

R

    C2H3O2-  (aq)   +       H2O(l)              HC2H3O2 (aq)     +         OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

iv. 0.044 L (0.10 mol OH-/1 L) = 0.0044 mol OH- added

pOH = - log [(0.0044 - 0.0032) mol OH- excess/(0.016 + 0.044) L total] = 1.70

pH = 14.00 – 1.70 = 12.30

 

 

Weak Base + Strong Acid Titration

 

When a weak base is titrated with a strong acid, the net ionic equation for the neutralization reaction is B (aq) + H+ (aq) BH+ (aq), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of acid needed to reach the equivalence point as follows:

 

LB × MB = nB = nH+/ MH+ = LH+

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial pH calculated using RICE chart for ionization of weak base B:

 

R

         B (aq)        +          H2O(l)                BH+ (aq)       +        OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

Point B = between start of titration and equivalence point = B/BH+ buffer solution created:

nH+ added = nB reacted = nBH+ produced

nB initial - nB reacted = nB excess

pOH = pKb + log (nBH+ produced/nB excess) = 14.00 - pH

halfway to equivalence point:  nBH+ produced = nB excess, so pOH = pKb = 14.00 - pH

 

Point C = equivalence point, pH calculated using RICE chart for hydrolysis of weak acid BH+:

 

R

      BH+ (aq)             H+ (aq)        +          B (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

pH < 7 at equivalence point

 

Point D = after equivalence point (hydrolysis of BH+ negligible):

pH = -log [H+]excess = -log [(nH+ - nB)/L total]

 

Sample Exercise 17W:


a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 54 mL of 0.10 M NH3 (Kb = 1.8 x 10-5) with 0.30 M HCl and calculate the volume of acid needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 11 mL of acid has been added

iii. the pH at the equivalence point

iv. the pH after 32 mL of acid has been added

 

Solution:


a. Cl- = spectator ion, neutralization reaction:  NH3 (aq) + H+ (aq) NH4+ (aq)

0.054 L (0.10 mol NH3/1 L) = 0.0054 mol NH3 = 0.0054 mol H+ (1 L/0.30 mol H+) = 0.018 L = 18 mL acid needed

 

b.

i.

R

        NH3 (aq)      +        H2O(l)         ⇌       NH4+ (aq)     +          OH- (aq)

I

0.10

 

0

0

C

-x

+x

+x

E

0.10 - x

x

x

 

 

ii.

0.011 L (0.30 mol H+/1 L) = 0.0033 mol H+ added = 0.0033 mol NH3 reacted = 0.0033 mol NH4+ produced

0.0054 mol NH3 initial - 0.0033 mol NH3 reacted = 0.0021 mol NH3 excess

pKb for NH3  = -log (1.8 x 10-5) = 4.74

pOH = 4.74 + log (0.0033 mol NH4+/0.0021 mol NH3) = 4.94

pH = 14.00 – 4.94 = 9.06

 

iii.

R

      NH4+ (aq)          H+ (aq)         +       NH3 (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

iv. 0.032 L (0.30 mol H+/1 L) = 0.0096 mol H+ added

pH = - log [(0.0096 - 0.0054) mol H+ excess/(0.054 + 0.032) L total] = 1.31

 

 

 

Section 17-13:  Acid-Base Indicators


One type of acid-base indicator is a weak acid where the neutral acid HInd has a different color than its conjugate base Ind-:

 

HInd (= color 1) H+ + Ind- (= color 2)

 

The pH of the solution being tested will determine the ratio of moles of Ind- to moles of HInd and, therefore, the color of the indicator as follows:

 

pH of solution = pKa for HInd + log (nInd-/nHInd)

 

pH << pKa:  nInd- <<  nHInd = color 1

pH = pKa:  nInd- = nHInd = mixture of color 1 and color 2

pH >> pKa:  nInd-  >>  nHInd = color 2

 

During each titration discussed above, the pH changes sharply by several units near the equivalence point.  An appropriate indicator for a titration will have one color before the equivalence point and then instantaneously change to a different color after the equivalence point due to the sharp change in pH.  The indicator chosen should have pKa for HInd close to the pH at the equivalence point:

 

titration:  choose indicator with pKa for HInd close to pH at equivalence point

 

Sample Exercise 17X:


Which indicator, thymolphthalein (Ka = 1 x 10-10) or bromcresol green (Ka = 8 x 10-6), would be the better choice for the titration in Sample Exercise 17W?

 

Solution:


pH at equivalence point = 5.19

pKa for thymolphthalein = -log (1 x 10-10) = 10.0

pKa for bromcresol green = -log (8 x 10-6) = 5.1

pKa for bromcresol green close to pH at equivalence point, so bromcresol green is better choice

 

 

 

Chapter 17 Practice Exercises and Review Quizzes:

 

17-1) A 74 mL sample of HI gas, measured at 55C and 694 mmHg, was dissolved in water to yield 48 mL of solution.  Calculate the molarity of hydrogen ion, the molarity of hydroxide ion, pH, and pOH in this solution.

Click for Solution

 

17-1) 

 

 

HI = strong acid:

R

       HI (aq)               H+ (aq)         +          I- (aq)     

I

0.052

0

0

C

-0.052

+0.052

+0.052

E

0

0.052

0.052

 

 

 

 

17-2) A 0.054 gram sample of barium hydroxide was dissolved in water to create 75 mL of solution.  Calculate the molarity of hydroxide ion, the molarity of hydrogen ion, pOH, and pH in this solution.

 

Click for Solution

 

17-2)

 

 

Ba(OH)2 = strong base:

 

R

  Ba(OH)2 (aq)  →      Ba2+ (aq)      +       2 OH- (aq)     

I

0.0042

0

0

C

-0.0042

+0.0042

+2(0.0042)

E

0

0.0042

0.0084

 

 

 

 

17-3) Write the acid ionization equation and calculate the pH and percent ionization of 0.87 M hypochlorous acid, HClO (Ka = 3.0 x 10-8).

 

Click for Solution

 

17-3)

R

      HClO (aq)          H+ (aq)        +        ClO- (aq)     

I

0.87

0

0

C

-x

+x

+x

E

0.87 - x

x

x

 

 

 

 

17-4) A 0.92 M hydrocyanic acid solution has a pH of 4.68.  Write the acid ionization equation and calculate Ka for hydrocyanic acid, HCN.

 

Click for Solution

 

17-4)

R

       HCN (aq)         H+ (aq)       +          CN- (aq)     

I

0.92

0

0

C

-x

+x

+x

E

0.92 - x

x

x

 

 

 

 

17-5) A 0.046 M solution of benzoic acid is 3.6% ionized.  Write the acid ionization equation and calculate the pH of the solution and Ka for benzoic acid, HC7H5O2.   

 

Click for Solution

 

17-5)

R

  HC7H5O2 (aq      H+ (aq)      +      C7H5O2- (aq)     

I

0.046

0

0

C

-x

+x

+x

E

0.046 - x

x

x

 

 

 

 

17-6) A nitrous acid, HNO2, solution has a pH of 2.58.  Given that Ka = 4.5 x 10-4 for nitrous acid, write the acid ionization equation and calculate the initial molarity of the nitrous acid solution.

  

Click for Solution

 

17-6)

R

     HNO2 (aq)         H+ (aq)     +           NO2- (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

 

17-7) Write the stepwise acid ionization equations and calculate the pH of 0.073 M sulfurous acid, H2SO3, which has the following acid ionization constants:

 

Ka1 = 1.7 x 10-2

Ka2 = 6.4 x 10-8

 

Click for Solution

 

17-7)

R

    H2SO3 (aq)          H+ (aq)       +        HSO3- (aq)     

I

0.073

0

0

C

-x

+x

+x

E

0.073 - x

x

x

 

 

 

2nd ionization:  HSO3- (aq) H+ (aq) + SO32- (aq)

    

 

17-8) Identify the Bronsted acids and bases in the forward and reverse directions for the reaction below:

 

HPO42- (aq) + H2AsO4- (aq) H2PO4- (aq) + HAsO42- (aq) 

 

Click for Solution

 

17-8)

forward reaction:  Bronsted acid = H2AsO4- (donates proton), Bronsted base = HPO42- (accepts proton)

reverse reaction:  Bronsted acid = H2PO4-, Bronsted base = HAsO42-

 

 

17-9) Write the formula for:

 

a. the conjugate acid of HC6H5O72-

b. the conjugate base of HC3H2O4-

 

Click for Solution

 

17-9)

a. conjugate acid = add H+ to formula = H2C6H5O7-

b. conjugate base = remove H+ from formula = C3H2O42-

 

 

17-10) Write the base ionization equation and calculate the pH and percent ionization of 0.11 M pyridine, C5H5N (Kb = 1.7 x 10-9).

 

Click for Solution

 

17-10)

R

     C5H5N (aq)     +         H2O(l)          C5H5NH+ (aq)    +       OH- (aq)

I

0.11

 

0

0

C

-x

+x

+x

E

0.11- x

x

x

 

 

 

 

17-11) A 0.72 M aniline solution has a pH of 9.26.  Write the base ionization equation and calculate Kb for aniline, C6H5NH2.

 

 

Click for Solution

 

17-11)

R

  C6H5NH2 (aq)    +        H2O(l)        ⇌   C6H5NH3+ (aq)   +       OH- (aq)

I

0.72

 

0

0

C

-x

+x

+x

E

0.72- x

x

x

 

 

 

 

 

17-12) A 0.052 M solution of ethylamine is 10.6% ionized.  Write the base ionization equation and calculate the pH of the solution and Kb for ethylamine, C2H5NH2.    

 

Click for Solution

 

17-12)

R

  C2H5NH2 (aq)    +        H2O(l)        C2H5NH3+ (aq)    +        OH- (aq)

I

0.052

 

0

0

C

-x

+x

+x

E

0.052- x

x

x

 

 

 

 

 

17-13) A trimethylamine, (CH3)3N, solution has a pH of 11.68.  Given that Kb = 6.4 x 10-5 for trimethylamine, write the base ionization equation and calculate the initial molarity of the trimethylamine solution.

 

Click for Solution

 

17-13)

R

   (CH3)3N (aq)    +          H2O(l)         (CH3)3NH+ (aq)   +       OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

 

 

17-14) Draw Lewis structures for arsenic acid, H3AsO4, and sulfuric acid, H2SO4.  Which is the stronger acid?  Give two reasons to justify your answer.

 

Click for Solution

 

17-14)

 

 

 

Sulfuric acid is stronger acid because:

 

1. H2SO4 has more terminal oxygens (4-2=2) than H3AsO4 (4-3=1).

2. Electronegativity for central atom S is higher than for central atom As.

 

 

17-15) Which of the two acids shown below is the stronger acid?  Give two reasons to justify your answer.

 


 

 

Click for Solution

 

17-15)

The top acid is stronger because:

 

1. Electronegativity of Cl is higher than I.

2. Cl atoms are closer to ionizable proton than I atoms.

 

 

17-16) Predict whether a solution of each compound below will be acidic, basic, or neutral.  For solutions that are not neutral, show all relevant hydrolysis reactions that affect the pH and also calculate the equilibrium constant for each reaction you write using information from the following data tables:

 

Acid

Ka

HClO

3.0 x 10-8

HF

6.8 x 10-4

 

Base

Kb

NH3

1.8 x 10-5

(CH3)3N

6.4 x 10-5

 

a. KF

b. (CH3)3NHClO [composed of (CH3)3NH+ and ClO-]

c. Ca(ClO4)2

d. NH4Br

 

 

Click for Solution

 

17-16)

a. K+ = spectator ion, F- hydrolyzes as a weak base = solution is basic:

 

F- (aq) + H2O (l) HF (aq) + OH- (aq)

 

 

b. Both ions hydrolyze, so we must compare equilibrium constants:

 

(CH3)3NH+ (aq) H+ (aq) + (CH3)3N (aq) 

 

 

ClO- (aq) + H2O (l) HClO (aq) + OH- (aq)

 

 

Ka < Kb = solution is basic

 

c. Ca2+ and ClO4- = spectator ions = solution is neutral

 

d. Br- = spectator ion, NH4+ hydrolyzes as weak acid = solution is acidic:

 

NH4+ (aq) H+ (aq) + NH3 (aq)

 

 

 

17-17) For each solution below, show any relevant hydrolysis reactions and calculate the pH.

 

a. 1.8 M NaNO2 (Ka for HNO2 = 4.5 x 10-4)

b. 0.66 M CH3NH3NO3 [composed of CH3NH3+ and NO3-] (Kb for CH3NH2 = 4.4 x 10-4)

 

 

Click for Solution

 

17-17)

a. Na+ = spectator ion, NO2- hydrolyzes as a weak base:

 

R

       NO2- (aq)    +          H2O(l)                HNO2 (aq)     +         OH- (aq)

I

1.8

 

0

0

C

-x

+x

+x

E

1.8- x

x

x

 

 

 

b. NO3- = spectator ion, CH3NH3+ hydrolyzes as a weak acid:

 

R

  CH3NH3+ (aq)     H+ (aq)       +      CH3NH2 (aq)     

I

0.66

0

0

C

-x

+x

+x

E

0.66 - x

x

x

 

 

 

 

17-18) Predict whether a solution of potassium hydrogen phosphate, K2HPO4, will be acidic or basic.  Show all relevant reactions that affect the pH and also give the value of the equilibrium constant for each reaction you write.  For phosphoric acid, H3PO4:

 

Ka1 = 7.5 x 10-3

Ka2 = 6.2 x 10-8

Ka3 = 4.2 x 10-13

 

Click for Solution

17-18)

K+ = spectator ion

 

weak acid ionization reaction: 

HPO42- (aq) H+ (aq) + PO43- (aq)    

Ka for HPO42- = Ka3 for H3PO4 = 4.2 x 10-13


weak base hydrolysis reaction:

HPO42- (aq) + H2O (l) H2PO4- (aq) + OH- (aq)    

Kb for HPO42- = 1.0 x 10-14/(Ka for H2PO4- = Ka2 for H3PO4 = 6.2 x 10-8) = 1.6 x 10-7

 

Ka < Kb = solution is basic

 

 

17-19) Will the reaction of Cl2O7 and water produce HClO3 or HClO4?  Write a balanced equation for the reaction.

 

Click for Solution

 

17-19) The reaction of Cl2O7 (oxidation number of Cl = +7) and water will produce HClO4 (oxidation number of Cl = +7) rather than HClO3 (oxidation number of Cl = +5).  The balanced equation will be Cl2O7 + H2O 2 HClO4.

 

 

17-20)

a. An unknown monoprotic weak acid was found to be 14.13% carbon, 1.78% hydrogen, and 74.67% iodine by mass, with the remainder being oxygen.  Determine the empirical formula of the acid.

b. In a separate experiment, 9.03 grams of the acid was dissolved in 35 mL of water and then titrated with 0.414 M calcium hydroxide.  The volume of base required to reach the equivalence point was 32.1 mL.  Calculate the molar mass and determine the molecular formula of the acid.

 

Click for Solution

 

17-20) a. 100% - 14.13% C – 1.78% H - 74.67% I = 9.42% O by mass

 

Assume one hundred grams of unknown acid:

 

 

1.177 mol C: 1.77 mol H: 0.5884 mol I: 0.5888 mol O (divide each by 0.5884)

= 2 mol C: 3 mol H: 1 mol I: 1 mol O

Therefore, empirical formula is C2H3IO.

 

b. 0.414 M Ca(OH)2 0.414 M Ca2+ and 2 x 0.414 M = 0.828 M OH-

HA (aq) + OH- (aq) H2O (l) + A- (aq)

 

 

Since the acid is monoprotic, we can rewrite the molecular formula as HC4H5I2O2.

 

17-21) Calculate the initial molarity of a potassium acetate, KC2H3O2, solution that has a pH of 8.91 given that Ka = 1.8 x 10-5 for acetic acid, HC2H3O2. 

Click for Solution

 

17-21) K+ = spectator ion, C2H3O2- hydrolyzes as a weak base:

 

R

   C2H3O2- (aq)    +         H2O(l)          ⇌      HC2H3O2 (aq)   +       OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

17-22) Calculate the initial molarity of a CH3NH3Cl [composed of CH3NH3+ and Cl-] solution that has a pH of 5.85 given that Kb = 4.4 x 10-4 for CH3NH2. 

Click for Solution

17-22) Cl- = spectator ion, CH3NH3+ hydrolyzes as a weak acid:

 

R

  CH3NH3+ (aq)      H+ (aq)       +     CH3NH2 (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

 

17-23) Given that a 1.6 M NaCN solution has a pH of 11.76, calculate Ka and pKa for HCN. 

Click for Solution

17-23) Na+ = spectator ion, CN- hydrolyzes as a weak base:

 

R

     CN- (aq)         +       H2O(l)                   HCN (aq)      +        OH- (aq)

I

1.6

 

0

0

C

-x

+x

+x

E

1.6 - x

x

x

 

 

 

 

17-24) Given that a 0.87 M C6H5NH3Br [composed of C6H5NH3+ and Br-] solution has a pH of 2.35, calculate Kb and pKb for C6H5NH2.

Click for Solution

17-24) Br- = spectator ion, C6H5NH3+ hydrolyzes as a weak acid:

 

R

 C6H5NH3+ (aq)         H+ (aq)      +     C6H5NH2 (aq)     

I

0.87

0

0

C

-x

+x

+x

E

0.87 - x

x

x

 

 

 

 

 

17-25) Calculate the pH of a solution containing 6.5 grams of KC2H3O2 in 50. mL of 0.88 M HC2H3O2 (Ka = 1.8 x 10-5).

Click for Solution

17-25)

K+ = spectator ion, HC2H3O2/C2H3O2- = buffer solution:

 

 

17-26) Calculate the pH of a solution containing 8.5 grams of CH3NH3Br [composed of CH3NH3+ and Br-]  in 88 mL of 0.92 M CH3NH2 (Kb = 4.4 x 10-4).

Click for Solution

17-26)

Br- = spectator ion, CH3NH2/CH3NH3+ = buffer solution:

 

 

17-27) Write the net ionic equation for the neutralization reaction that occurs and calculate the pH when:

 

a. 0.003 mol HNO3 is added to the solution in Practice Exercise 17-25

b. 0.003 mol KOH is added to the solution in Practice Exercise 17-25

c. 0.003 mol HNO3 is added to the solution in Practice Exercise 17-26

d. 0.003 mol KOH is added to the solution in Practice Exercise 17-26

Click for Solution

17-27)

a. NO3- = spectator ion

neutralization reaction:  C2H3O2- (aq) + H+ (aq) HC2H3O2 (aq)

pH = 4.74 + log (0.066 – 0.003) mol C2H3O2-/(0.044 + 0.003) mol HC2H3O2 = 4.87 

 

b. K+ = spectator ion

neutralization reaction: HC2H3O2 (aq) + OH- (aq) C2H3O2- (aq) + H2O (l)

pH = 4.74 + log (0.066 + 0.003) mol C2H3O2-/(0.044 - 0.003) mol HC2H3O2 = 4.97

 

c. NO3- = spectator ion

neutralization reaction:  CH3NH2 (aq) + H+ (aq) CH3NH3+ (aq)

pOH = 3.36 + log (0.076 + 0.003) mol CH3NH3+/(0.081 - 0.003) mol CH3NH2 = 3.37

pH = 14.00 – 3.37 = 10.63 

 

d. K+ = spectator ion

neutralization reaction: CH3NH3+ (aq) + OH- (aq) CH3NH2 (aq) + H2O (l)

pOH = 3.36 + log (0.076 - 0.003) mol CH3NH3+/(0.081 + 0.003) mol CH3NH2 = 3.30

pH = 14.00 – 3.30 = 10.70

 

 

17-28) Write the net ionic equation for the neutralization reaction that occurs in each aqueous mixture below.  Which one of the four reactions creates a buffer solution?  For each mixture, describe the method of calculating the pH of the resulting solution after the neutralization reaction is complete.

 

a. 0.5 mol KOH + 0.5 mol NH4I

b. 0.2 mol HClO4 + 0.5 mol NaC7H5O2

c. 0.2 mol HClO + 0.5 mol NaOH

d. 0.5 mol HNO3 + 0.2 mol C5H5N (a weak base)

Click for Solution

17-28)

a. K+ and I- = spectator ions

neutralization reaction:  NH4+ (aq) + OH- (aq) NH3 (aq) + H2O (l)

n of NH4+ = nOH-:  NH3 solution created, pOH = 14.00 - pH calculated using RICE chart for ionization of weak base NH3

 

b. ClO4- and Na+ = spectator ions

neutralization reaction:  C7H5O2- (aq) + H+ (aq) HC7H5O2 (aq)

n of C7H5O2- > nH+:  HC7H5O2/ C7H5O2- buffer created, pH = pKa + log (n of C7H5O2- excess/n of HC7H5O2 produced)

 

c. Na+ = spectator ion

neutralization reaction:  HClO (aq) + OH- (aq) ClO- (aq) + H2O (l)

nHClO < nOH-:  OH-/ClO- solution created, pOH = 14.00 – pH = -log [OH-]excess (hydrolysis of ClO- negligible)

 

d. NO3- = spectator ion

neutralization reaction:  C5H5N (aq) + H+ (aq) C5H5NH+ (aq)

n of C5H5N < nH+:  H+/ C5H5NH+ solution created, pH = -log [H+]excess (hydrolysis of C5H5NH+ negligible)

 

 

17-29) a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 42 mL of 0.15 M KOH with 0.30 M HI and calculate the volume of acid needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 13 mL of acid has been added

iii. the pH at the equivalence point

iv. the pH after 25 mL of acid has been added

Click for Solution

17-29)

a. K+ and I- = spectator ions, neutralization reaction:  OH- (aq) + H+ (aq) H2O (l)

0.042 L (0.15 mol OH-/1 L) = 0.0063 mol OH- = 0.0063 mol H+ (1 L/0.30 mol H+) = 0.021 L = 21 mL acid needed

 

b.

i. pOH = -log (0.15) = 0.82

pH = 14.00 - 0.82 = 13.18

 

ii. 0.013 L (0.30 mol H+/1 L) = 0.0039 mol H+ added

pOH = -log [(0.0063 - 0.0039) mol OH- excess/(0.042 + 0.013) L total] = 1.36

pH = 14.00 - 1.36 = 12.64

 

iii. only H2O present, pH = 7

 

iv. 0.025 L (0.30 mol H+/1 L) = 0.0075 mol H+ added

pH = - log [(0.0075 - 0.0063) mol H+ excess/(0.042 + 0.025) L total] = 1.75

 

 

17-30)

a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 36 mL of 0.20 M HNO2 (Ka = 4.5 x 10-4) with 0.30 M NaOH and calculate the volume of base needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 11 mL of base has been added

iii. the pH at the equivalence point

iv. the pH after 33 mL of base has been added

Click for Solution

17-30)

a. Na+ = spectator ion, neutralization reaction:  HNO2 (aq) + OH- (aq) H2O (l) + NO2- (aq)

0.036 L (0.20 mol HNO2/1 L) = 0.0072 mol HNO2 = 0.0072 mol OH- (1 L/0.30 mol OH-) = 0.024 L = 24 mL base needed

 

b.

i.

R

    HNO2 (aq)              H+ (aq)         +      NO2- (aq)     

I

0.20

0

0

C

-x

+x

+x

E

0.20 - x

x

x

 

 

ii.

0.011 L (0.30 mol OH-/1 L) = 0.0033 mol OH- added = 0.0033 mol HNO2 reacted = 0.0033 mol NO2- produced

0.0072 mol HNO2 initial - 0.0033 mol HNO2 reacted = 0.0039 mol HNO2 excess

pKa for HNO2  = -log (4.5 x 10-4) = 3.35

pH = 3.35 + log (0.0033 mol NO2-/0.0039 mol HNO2) = 3.28

 

iii.

R

     NO2-  (aq)       +         H2O(l)               HNO2 (aq)     +        OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

iv. 0.033 L (0.30 mol OH-/1 L) = 0.0099 mol OH- added

pOH = - log [(0.0099 - 0.0072) mol OH- excess/(0.036 + 0.033) L total] = 1.41

pH = 14.00 – 1.41 = 12.59

 

 

17-31)

a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 18 mL of 0.30 M CH3NH2 (Kb = 4.4 x 10-4) with 0.20 M HBr and calculate the volume of acid needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 16 mL of acid has been added

iii. the pH at the equivalence point

iv. the pH after 38 mL of acid has been added

Click for Solution

17-31)

a. Br- = spectator ion, neutralization reaction:  CH3NH2 (aq) + H+ (aq) CH3NH3+ (aq)

0.018 L (0.30 mol NH3/1 L) = 0.0054 mol CH3NH2 = 0.0054 mol H+ (1 L/0.20 mol H+) = 0.027 L = 27 mL acid needed

 

b.

i.

R

   CH3NH2 (aq)     +       H2O(l)             CH3NH3+ (aq)   +       OH- (aq)

I

0.30

 

0

0

C

-x

+x

+x

E

0.30 - x

x

x

 

 

ii.

0.016 L (0.20 mol H+/1 L) = 0.0032 mol H+ added = 0.0032 mol CH3NH2 reacted = 0.0032 mol CH3NH3+ produced

0.0054 mol CH3NH2 initial - 0.0032 mol CH3NH2 reacted = 0.0022 mol CH3NH2 excess

pKb for CH3NH2  = -log (4.4 x 10-4) = 3.36

pOH = 3.36 + log (0.0032 mol CH3NH3+/0.0022 mol CH3NH2) = 3.52

pH = 14.00 – 3.52 = 10.48

 

iii.

R

  CH3NH3+ (aq)       H+ (aq)       +    CH3NH2 (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

iv. 0.038 L (0.20 mol H+/1 L) = 0.0076 mol H+ added

pH = - log [(0.0076 - 0.0054) mol H+ excess/(0.018 + 0.038) L total] = 1.41

 

 

17-32) Which indicator, metacresol purple (Ka = 5 x 10-9) or 2,4-dinitrophenol (Ka = 1 x 10-4), would be the better choice for the titration in Practice Exercise 17-30?

Click for Solution

17-32)

pH at equivalence point = 8.20

pKa for metacresol purple = -log (5 x 10-9) = 8.3

pKa for 2,4-dinitrophenol = -log (1 x 10-4) = 4.0

pKa for metacresol purple close to pH at equivalence point, so metacresol purple is better choice

 

 

Click for Review Quiz 1

Click for Review Quiz 1 Answers